# 16.7 Damped harmonic motion  (Page 2/5)

 Page 2 / 5

## Damping an oscillatory motion: friction on an object connected to a spring

Damping oscillatory motion is important in many systems, and the ability to control the damping is even more so. This is generally attained using non-conservative forces such as the friction between surfaces, and viscosity for objects moving through fluids. The following example considers friction. Suppose a 0.200-kg object is connected to a spring as shown in [link] , but there is simple friction between the object and the surface, and the coefficient of friction ${\mu }_{k}$ is equal to 0.0800. (a) What is the frictional force between the surfaces? (b) What total distance does the object travel if it is released 0.100 m from equilibrium, starting at $v=0$ ? The force constant of the spring is $k=\text{50}\text{.}0 N/m\text{}$ .

Strategy

This problem requires you to integrate your knowledge of various concepts regarding waves, oscillations, and damping. To solve an integrated concept problem, you must first identify the physical principles involved. Part (a) is about the frictional force. This is a topic involving the application of Newton’s Laws. Part (b) requires an understanding of work and conservation of energy, as well as some understanding of horizontal oscillatory systems.

Now that we have identified the principles we must apply in order to solve the problems, we need to identify the knowns and unknowns for each part of the question, as well as the quantity that is constant in Part (a) and Part (b) of the question.

Solution a

1. Choose the proper equation: Friction is $f={\mu }_{k}\text{mg}$ .
2. Identify the known values.
3. Enter the known values into the equation:
$f=\text{(0.0800)}\left(0\text{.200 kg)}\left(9\text{.80 m}/{\text{s}}^{\text{2}}\text{)}.$
4. Calculate and convert units: $f=\text{0.157 N}.$

Discussion a

The force here is small because the system and the coefficients are small.

Solution b

Identify the known:

• The system involves elastic potential energy as the spring compresses and expands, friction that is related to the work done, and the kinetic energy as the body speeds up and slows down.
• Energy is not conserved as the mass oscillates because friction is a non-conservative force.
• The motion is horizontal, so gravitational potential energy does not need to be considered.
• Because the motion starts from rest, the energy in the system is initially ${\text{PE}}_{\mathrm{el,i}}=\left(1/2\right){\text{kX}}^{2}$ . This energy is removed by work done by friction ${W}_{\text{nc}}=–\text{fd}$ , where $d$ is the total distance traveled and $f={\mu }_{\text{k}}\text{mg}$ is the force of friction. When the system stops moving, the friction force will balance the force exerted by the spring, so ${\text{PE}}_{\text{e1,f}}=\left(1/2\right){\text{kx}}^{2}$ where $x$ is the final position and is given by
$\begin{array}{lll}{F}_{\text{el}}& =& f\\ \text{kx}& =& {\mu }_{\text{k}}\text{mg}\\ x& =& \frac{{\mu }_{\text{k}}\text{mg}}{k}\end{array}.$
1. By equating the work done to the energy removed, solve for the distance $d$ .
2. The work done by the non-conservative forces equals the initial, stored elastic potential energy. Identify the correct equation to use:
${\text{W}}_{\text{nc}}=\Delta \left(\text{KE}+\text{PE}\right)={\text{PE}}_{\text{el,f}}-{\text{PE}}_{\text{el,i}}=\frac{1}{2}k\left({\left(\frac{{\mu }_{k}\mathit{\text{mg}}}{k}\right)}^{2}-{X}^{2}\right).$
3. Recall that ${W}_{\text{nc}}=–\text{fd}$ .
4. Enter the friction as $f={\mu }_{\text{k}}\text{mg}$ into ${W}_{\text{nc}}=–\text{fd}$ , thus
${W}_{\text{nc}}={–\mu }_{\text{k}}\text{mgd}.$
5. Combine these two equations to find
$\frac{1}{2}k\left({\left(\frac{{\mu }_{k}\text{mg}}{k}\right)}^{2}-{X}^{2}\right)=-{\mu }_{\text{k}}\text{mgd}.$
6. Solve the equation for $d$ :
$d=\frac{\text{k}}{{\text{2}\mu }_{\text{k}}\text{mg}}\left({X}^{2}–{\left(\frac{{\mu }_{\text{k}}\text{mg}}{k}\right)}^{2}\right).$
7. Enter the known values into the resulting equation:
$d=\frac{\text{50}\text{.}0 N/m\text{}}{2\left(0\text{.}\text{0800}\right)\left(0\text{.}\text{200}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)}\left({\left(0\text{.}\text{100}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}^{2}-{\left(\frac{\left(0\text{.}\text{0800}\right)\left(0\text{.}\text{200}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)}{\text{50}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{N/m}}\right)}^{2}\right).$
8. Calculate $d$ and convert units:
$d=1\text{.}\text{59}\phantom{\rule{0.25em}{0ex}}\text{m}.$

Discussion b

This is the total distance traveled back and forth across $x=0$ , which is the undamped equilibrium position. The number of oscillations about the equilibrium position will be more than $d/X=\left(1\text{.}\text{59}\phantom{\rule{0.25em}{0ex}}\text{m}\right)/\left(0\text{.}\text{100}\phantom{\rule{0.25em}{0ex}}\text{m}\right)=\text{15}\text{.}9$ because the amplitude of the oscillations is decreasing with time. At the end of the motion, this system will not return to $x=0$ for this type of damping force, because static friction will exceed the restoring force. This system is underdamped. In contrast, an overdamped system with a simple constant damping force would not cross the equilibrium position $x=0$ a single time. For example, if this system had a damping force 20 times greater, it would only move 0.0484 m toward the equilibrium position from its original 0.100-m position.

This worked example illustrates how to apply problem-solving strategies to situations that integrate the different concepts you have learned. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknowns using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life.

Why are completely undamped harmonic oscillators so rare?

Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator.

Describe the difference between overdamping, underdamping, and critical damping.

An overdamped system moves slowly toward equilibrium. An underdamped system moves quickly to equilibrium, but will oscillate about the equilibrium point as it does so. A critically damped system moves as quickly as possible toward equilibrium without oscillating about the equilibrium.

## Section summary

• Damped harmonic oscillators have non-conservative forces that dissipate their energy.
• Critical damping returns the system to equilibrium as fast as possible without overshooting.
• An underdamped system will oscillate through the equilibrium position.
• An overdamped system moves more slowly toward equilibrium than one that is critically damped.

## Conceptual questions

Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.)

How would a car bounce after a bump under each of these conditions?

• overdamping
• underdamping
• critical damping

Most harmonic oscillators are damped and, if undriven, eventually come to a stop. How is this observation related to the second law of thermodynamics?

## Problems&Exercises

The amplitude of a lightly damped oscillator decreases by $3\text{.}0%\text{}$ during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

#### Questions & Answers

is there a formula for magnitudes of displacement
oyeronke Reply
A Body of maas m slides down an incline and reached the bottoms with a velocity v
Aditya Reply
What is mass
Aditya
mass is the amount of matter a body contains.
prakash
what is model
Chisom Reply
what is a dimension
John
When using the Conservation of Energy equation, do we substitute the energy as a negative quantity when the energies on a single object are exerting forces opposite to one another?
Jennifer Reply
Ex. On an inclined plane, gravitational potential energy, friction energy/work and spring potential energy. (Let's say that the spring is keeping the box from sliding down the slope.) How do we use this in the equation? I'm so confused
Jennifer
Oh! And if there's kinetic energy that is exerting a force opposite to the spring, what do we do?
Jennifer
Then we subtract the k.e. from force exerted from newton's 2nd law.
Prem
Subtract energy from force? They're different units
Jennifer
why is it dat when using double pan balance the known and unknown mass are the same
Victor Reply
discuss the uses of energy in the following sectors of economy security and education
amajuoyi
why the current produce during dc motot is not use for its working rather we have to supply current outside
Tanveer
is there more then 4 dimensions
Miguel Reply
hii
princy
hi
Miguel
hello I kinda need help in physics... a lot
Brown
Brown. what kind of help
Jeff
when it comes to physics stick with the basics don't overthink things
Jeff
yes
ayesha
sticking to the basics will take you farther than overwhelming yourself with more than you need to physics is simple keep it simple
Jeff
thk u Ayesha
Jeff
for real....? so I've got to know the fundamentals and use the formula to solve any problem
Brown
read Stephan hawkings a brief history of time
ayesha
it'll help you understand better than summing up formulas or ready textbooks
ayesha
physics isn't hard it's just understanding and applying the formulas if u need help ask any question
ayesha
okay...because I've got an exam next year February a Computer based exam
Brown
start with a pace a plan and stick to it
ayesha
well best of luck can't help you much there contact your teachers for tips and helpful notes
ayesha
hi
Varun
hello
Ibrahim
yeah
oyeronke
how can we find absolute uncertainty
ayesha Reply
it what?
Luke
in physics
ayesha
the basic formula is uncertainty in momentum multiplied buy uncertainty In position is greater than or equal to 4×pi/2. same formula for energy and time
Luke
I have this one question can you please look it up it's 9702/22/O/N/17 Question 1 B 3
ayesha
what
uma
would you like physics?
Suthar
yes
farooq
precision or absolute uncertainty is always equal to least count of that instrument
Iram
how do I unlock the MCQ and the Essay?
Ojeh Reply
what is the dimension of strain
Joy Reply
Is there a formula for time of free fall given that the body has initial velocity? In other words, formula for time that takes a downward-shot projectile to hit the ground. Thanks!
Cyclone Reply
hi
Agboro
hiii
Chandan
Hi
Sahim
hi
Jeff
hey
Priscilla
sup guys
Bile
Hy
Kulsum
What is unit of watt?
Kulsum
watt is the unit of power
Rahul
p=f.v
Rahul
watt can also be expressed as Nm/s
Rahul
what s i unit of mass
Maxamed
SI unit of mass is Kg(kilogram).
Robel
what is formula of distance
Maxamed
Formula for for the falling body with initial velocity is:v^2=v(initial)^2+2*g*h
Mateo
i can't understand
Maxamed
we can't do this calculation without knowing the height of the initial position of the particle
Chathu
sorry but no more in science
Imoreh
2 forces whose resultant is 100N, are at right angle to each other .if one of them makes an angle of 30 degree with the resultant determine it's magnitude
Victor Reply
50 N... (50 *1.732)N
Sahim
Plz cheak the ans and give reply..
Sahim
50 N...(50 *1.732)N
Ibrahim
show the working
usiomon
what is the value of f1 and f2
Syed
what is the value of force 1 and force 2.
Syed
.
muhammad
Is earth is an inertial frame?
Sahim Reply
The abacus (plural abaci or abacuses), also called a counting frame, is a calculating tool that was in use in Europe, China and Russia, centuries before the adoption of the written Hindu–Arabic numeral system
Sahim
thanks
Irungu
Most welcome
Sahim
Hey.. I've a question.
Sahim Reply
Is earth inertia frame?
Sahim
only the center
Shii
What is an abucus?
Irungu
what would be the correct interrogation "what is time?" or "how much has your watch ticked?"
prakash Reply
someone please give answer to this.
prakash
hmmmm
oyeronke
coming
oyeronke
time is the distance measured on clock
KINGSFORD
time is that in which events are distinguishable with reference to before and after
Omo
time measure of duration interval of period
Sewmeho
a load of 20N on a wire of cross sectional area 8×10^-7m produces an extension of 10.4m. calculate the young modules of the material of the wire is of length 5m
Ebenezer Reply
Young's modulus = stress/strain strain = extension/length (x/l) stress = force/area (F/A) stress/strain is F l/A x
El
so solve it
Ebenezer
please
Ebenezer

### Read also:

#### Get the best College physics course in your pocket!

Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

 By By By Mldelatte By Qqq Qqq